Clarifier

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

9 formulas

10 exam-style examples

~60 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Clarifier within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what clarifier describes physically and when it applies.
State every one of the 9 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Clarifier is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: clarifier.
Capstone Studio instructional photograph

Environmental Engineering — Clarifier: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 9 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Overflow rate	Quantity produced by "Overflow rate = Hydraulic loading rate = vo = Q/Asurface" — read its definition and unit from the handbook line directly above the equation.
vo	Quantity produced by "vo = critical settling velocity; terminal settling velocity of smallest particle that is 100% removed" — read its definition and unit from the handbook line directly above the equation.
Weir loading	Quantity produced by "Weir loading = weir overflow rate, WOR = Q/Weir Length" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = flowrate" — read its definition and unit from the handbook line directly above the equation.
Ax	Quantity produced by "Ax = cross-sectional area" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = surface area, plan view" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = tank volume" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where
Typical Primary Clarifier Efficiency Percent Removal
Overflow rates
1,200 1,000 800 600
(gpd/ft2) (gpd/ft2) (gpd/ft2) (gpd/ft2)
48.9 40.7 32.6 24.4
(m/d) (m/d) (m/d) (m/d)
Suspended
54% 58% 64% 68%
Solids
BOD5 30% 32% 34% 36%
Design Criteria for Sedimentation Basins
Hydraulic
Overflow Rate Solids Loading Rate
Residence Depth
Type of Basin Average Peak Average Peak Time (ft)
(gpd/ft2) (m3/m2•d) (gpd/ft2) (m3/m2•d) (lb/ft2−d) (kg/m2•h) (lb/ft2−h) (kg/m2•h) (hr)

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydraulic detention time in a tank — Clarifier

A treatment tank holds 24,702 gal and treats 5.9 MGD. Find the hydraulic detention time.

Given

V = 24,702 gal
Q = 5.9 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$5.9 \times 1 0^{6} = 5, 900, 000 g a l / d a y$
Substituting
$θ = 24702/5, 900, 000 = 0.0042 d a y$
Convert
$θ = 0.0042 (24) = 0.10 h r$

Answer: θ ≈ 0.10 hr

Why the other options are there

0.004 hr (day/hour conversion missed)
238.8 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 2

Circular clarifier overflow rate, detention time and Stokes settling — Clarifier

A circular clarifier 24 m in diameter and 3.5 m deep treats 23,156 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 40 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 23,156 m³/d
D = 24 m, depth = 3.5 m
d_p = 40 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (24)^{2} = 452.4 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 23156/452.4 = 51.19 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 1, 583/23156 = 1.64 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 23156/(π × 24) = 307.1 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (4.0 e - 5)^{2} / (18 \times 0.001) = 124.3 m / d$
Compare
$v_{s} = 124.3 m / d > S O R = 51.19 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 51.2 m/d, τ = 1.6 h, weir loading = 307.1 m³/m·d, v_s = 124.3 m/d

Why the other options are there

SOR = 87.7 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 3

Hydraulic detention time in a tank — Clarifier (2)

A treatment tank holds 25,661 gal and treats 5.4 MGD. Find the hydraulic detention time.

Given

V = 25,661 gal
Q = 5.4 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$5.4 \times 1 0^{6} = 5, 400, 000 g a l / d a y$
Substituting
$θ = 25661/5, 400, 000 = 0.0048 d a y$
Convert
$θ = 0.0048 (24) = 0.11 h r$

Answer: θ ≈ 0.11 hr

Why the other options are there

0.005 hr (day/hour conversion missed)
210.4 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 4

Circular clarifier overflow rate, detention time and Stokes settling — Clarifier (2)

A circular clarifier 33 m in diameter and 3.5 m deep treats 9,669 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 70 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 9,669 m³/d
D = 33 m, depth = 3.5 m
d_p = 70 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (33)^{2} = 855.3 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 9669/855.3 = 11.30 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 994/9669 = 7.43 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 9669/(π × 33) = 93.3 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (7.0 e - 5)^{2} / (18 \times 0.001) = 380.7 m / d$
Compare
$v_{s} = 380.7 m / d > S O R = 11.30 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 11.3 m/d, τ = 7.4 h, weir loading = 93 m³/m·d, v_s = 380.7 m/d

Why the other options are there

SOR = 26.6 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 5

Hydraulic detention time in a tank — Clarifier (3)

A treatment tank holds 43,208 gal and treats 3.8 MGD. Find the hydraulic detention time.

Given

V = 43,208 gal
Q = 3.8 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$3.8 \times 1 0^{6} = 3, 800, 000 g a l / d a y$
Substituting
$θ = 43208/3, 800, 000 = 0.0114 d a y$
Convert
$θ = 0.0114 (24) = 0.27 h r$

Answer: θ ≈ 0.27 hr

Why the other options are there

0.011 hr (day/hour conversion missed)
87.95 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 6

Circular clarifier overflow rate, detention time and Stokes settling — Clarifier (3)

A circular clarifier 36 m in diameter and 3.0 m deep treats 11,696 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 70 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 11,696 m³/d
D = 36 m, depth = 3.0 m
d_p = 70 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (36)^{2} = 1, 018 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 11696/1,018 = 11.49 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 3, 054/11696 = 6.27 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 11696/(π × 36) = 103.4 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (7.0 e - 5)^{2} / (18 \times 0.001) = 380.7 m / d$
Compare
$v_{s} = 380.7 m / d > S O R = 11.49 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 11.5 m/d, τ = 6.3 h, weir loading = 103.4 m³/m·d, v_s = 380.7 m/d

Why the other options are there

SOR = 34.5 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 7

Hydraulic detention time in a tank — Clarifier (4)

A treatment tank holds 174,322 gal and treats 3.9 MGD. Find the hydraulic detention time.

Given

V = 174,322 gal
Q = 3.9 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$3.9 \times 1 0^{6} = 3, 900, 000 g a l / d a y$
Substituting
$θ = 174322/3, 900, 000 = 0.0447 d a y$
Convert
$θ = 0.0447 (24) = 1.07 h r$

Answer: θ ≈ 1.07 hr

Why the other options are there

0.045 hr (day/hour conversion missed)
22.37 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 8

Circular clarifier overflow rate, detention time and Stokes settling — Clarifier (4)

A circular clarifier 39 m in diameter and 4.0 m deep treats 25,852 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 20 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 25,852 m³/d
D = 39 m, depth = 4.0 m
d_p = 20 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (39)^{2} = 1, 195 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 25852/1,195 = 21.64 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 4, 778/25852 = 4.44 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 25852/(π × 39) = 211.0 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (2.0 e - 5)^{2} / (18 \times 0.001) = 31.08 m / d$
Compare
$v_{s} = 31.08 m / d > S O R = 21.64 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 21.6 m/d, τ = 4.4 h, weir loading = 211.0 m³/m·d, v_s = 31.1 m/d

Why the other options are there

SOR = 52.7 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 9

Hydraulic detention time in a tank — Clarifier (5)

A treatment tank holds 60,703 gal and treats 1.1 MGD. Find the hydraulic detention time.

Given

V = 60,703 gal
Q = 1.1 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$1.1 \times 1 0^{6} = 1, 100, 000 g a l / d a y$
Substituting
$θ = 60703/1, 100, 000 = 0.0552 d a y$
Convert
$θ = 0.0552 (24) = 1.32 h r$

Answer: θ ≈ 1.32 hr

Why the other options are there

0.055 hr (day/hour conversion missed)
18.12 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Example 10

Circular clarifier overflow rate, detention time and Stokes settling — Clarifier (5)

A circular clarifier 26 m in diameter and 5.0 m deep treats 17,946 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 80 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 17,946 m³/d
D = 26 m, depth = 5.0 m
d_p = 80 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (26)^{2} = 530.9 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 17946/530.9 = 33.80 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 655/17946 = 3.55 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 17946/(π × 26) = 219.7 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (8.0 e - 5)^{2} / (18 \times 0.001) = 497.2 m / d$
Compare
$v_{s} = 497.2 m / d > S O R = 33.80 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 33.8 m/d, τ = 3.6 h, weir loading = 219.7 m³/m·d, v_s = 497.2 m/d

Why the other options are there

SOR = 43.9 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Clarifier

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Clarifier contains 9 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Clarifier

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Overflowrate=Hydraulicloadingrate=vo=Q/Asurface

vo=criticalsettlingvelocity;terminalsettlingvelocityofsmallestparticlethatis100

Weirloading=weiroverflowrate,WOR=Q/WeirLength

Horizontalvelocity=approachvelocity=vh=Q/Across−section=Q/Ax

Hydraulicresidencetime=V/Q=θ

Q=flowrate

Ax=cross−sectionalarea

A=surfacearea,planview

V=tankvolume

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section